Question

If the crystal field splitting energy of [Co(NH$_3$)$_6$]$^{2+}$ is 5900 cm$^{-1}$, then the magnitude of its crystal field stabilization energy, in kJ mol$^{-1}$ (rounded off to one decimal place), is ________.

Hint:

For octahedral complexes, each $t_{2g}$ electron contributes $-0.4\Delta_0$, and each $e_g$ electron contributes $+0.6\Delta_0$.

Answer 1

Correct Answer: 84.2

Step 1: Recall CFSE for octahedral $d^7$ (high-spin) complex.
For $d^7$ high-spin configuration (in octahedral field): \[ \text{CFSE} = (3 \times -0.4\Delta_0) + (4 \times 0.6\Delta_0) = +0.6\Delta_0 \] Step 2: Convert $\Delta_0$ to energy.
\[ \Delta_0 = 5900 \, \text{cm}^{-1} = 5900 \times 1.986 \times 10^{-23} \, \text{J} \] \[ \Delta_0 = 1.171 \times 10^{-19} \, \text{J per ion} \] \[ \text{CFSE per mole} = 0.6 \times (1.171 \times 10^{-19} \times 6.022 \times 10^{23}) = 4.23 \times 10^{4} \, \text{J mol}^{-1} \] \[ \text{CFSE} = 42.3 \, \text{kJ mol}^{-1} \] However, the complex [Co(NH$_3$)$_6$]$^{2+}$ is low-spin ($d^7$), so \[ \text{CFSE} = (6 \times -0.4\Delta_0) + (1 \times 0.6\Delta_0) = -1.8\Delta_0 \] \[ = 1.8 \times 5900 \, \text{cm}^{-1} \times 1.986 \times 10^{-23} \times 6.022 \times 10^{23} = 141.5 \, \text{kJ mol}^{-1} \] Step 3: Conclusion.
CFSE = 141.5 kJ mol$^{-1}$.